Lemma 11.4.3. Let $V$ be a $k$ vector space. Let $K$ be a central $k$-algebra which is a skew field. Let $W \subset V \otimes _ k K$ be a two-sided $K$-sub vector space. Then $W$ is generated as a left $K$-vector space by $W \cap (V \otimes 1)$.

**Proof.**
Let $V' \subset V$ be the $k$-sub vector space generated by $v \in V$ such that $v \otimes 1 \in W$. Then $V' \otimes _ k K \subset W$ and we have

If $\overline{v} \in V/V'$ is a nonzero vector such that $\overline{v} \otimes 1$ is contained in $W/V' \otimes _ k K$, then we see that $v \otimes 1 \in W$ where $v \in V$ lifts $\overline{v}$. This contradicts our construction of $V'$. Hence we may replace $V$ by $V/V'$ and $W$ by $W/V' \otimes _ k K$ and it suffices to prove that $W \cap (V \otimes 1)$ is nonzero if $W$ is nonzero.

To see this let $w \in W$ be a nonzero element which can be written as $w = \sum _{i = 1, \ldots , n} v_ i \otimes k_ i$ with $n$ minimal. We may right multiply with $k_1^{-1}$ and assume that $k_1 = 1$. If $n = 1$, then we win because $v_1 \otimes 1 \in W$. If $n > 1$, then we see that for any $c \in K$

and hence $c k_ i - k_ i c = 0$ by minimality of $n$. This implies that $k_ i$ is in the center of $K$ which is $k$ by assumption. Hence $w = (v_1 + \sum k_ i v_ i) \otimes 1$ contradicting the minimality of $n$. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: